Review for Differential Calculus: Mathematics 102
Review problems by Leah Edelstein-Keshet: All rights reserved
University of British Columbia
November 29, 2013
ii
Leah Edelstein-Keshet
Exercises
0.1. Multiple Choice:
1 : The equation of the tangent line to the function y = f (x) at the point x0 is
(a) y = f ′ (x0 ) + f (x0 )(x − x0 )
(b) y = x0 + f (x0 )/f ′ (x0 )
(c) y = f (x) − f ′ (x)(x − x0 )
(d) y = f (x0 ) + f ′ (x0 )(x − x0 )
(e) y = f (x0 ) − f ′ (x0 )(x − x0 )
2 : The functions f (x) = x2 and g(x) = x3 are equal at x = 0 and at x = 1.
Between x = 0 and at x = 1, for what value of x are their graphs furthest
apart?
(a) x = 1/2 (b) x = 2/3 (c) x = 1/3 (d) x = 1/4 (e) x = 3/4
3 : Consider a point in the first quadrant on the hyperbola x2 − y 2 = 1 with
x = 2. The slope of the tangent line at that point is
√
√
√
√
(a) 2/ 3 (b) 2/ 5 (c) 1/ 3 (d) 5/2 (e) 2/3
4 : For a, b > 0, solving the equation ln(x) = 2 ln(a) − 3 ln(b) for x leads to
(a) x = e2a−3b (b) x = 2a − 3b (c) x = a2 /b3 (d) x = a2 b3 (e) x = (a/b)6
5 : The function y = f (x) = arctan(x) − (x/2) has local maXima (LX), local
minima (LM) and inflection points(IP) as follows:
(a) LX: x = 1, LM: x = −1, IP: x = 0.
(b) LX: x = −1, LM: x = 1, IP: x = 0.
(c) LX: x = −1, LM: x = 1, IP: none
√
√
(d) LX: x = 3, LM: x = − 3, IP: x = 0.
√
√
(e) LX: x = − 3, LM: x = 3, IP: x = 0.
0.2. Related Rates: Two spherical balloons are connected so that one inflates as the
other deflates, the sum of their volumes remaining constant. When the first balloon
has radius 10 cm and its radius is increasing at 3 cm/sec, the second balloon has
radius 20 cm. What is the rate of change of the radius of the second balloon? [The
volume of a sphere of radius r is V = (4/3)πr3 ].
0.3. Particle velocity: A particle is moving along the x axis so that its distance from the
origin at time t is given by
x(t) = (t + 2)3 + λt
where λ is a constant
(a) Determine the velocity v(t) and the acceleration a(t).
(b) Determine the minimum velocity over all time.
0.4. Motion: A particle’s motion is described by y(t) = t3 − 6t2 + 9t
where y(t) is the displacement (in metres) t is time (in seconds) and 0 ≤ t ≤ 4
seconds.
iii
iv
Review Problems
(a) During this time interval, when is the particle furthest from its initial position
?
(b) During this time interval, what is the greatest speed of the particle?
(c) What is the total distance (including both forward and backward directions)
that the particle has travelled during this time interval?
0.5. Fish generations: In Fish River, the number of salmon (in thousands), x, in a given
year is linked to the number of salmon (in thousands), y, in the following year by
the function
y = Axe−bx
where A, b > 0 are constants.
(a) For what number of salmon is there no change in the number from one year to
the next?
(b) Find the number of salmon that would yield the largest number of salmon in
the following year.
0.6. Blood alcohol: Blood alcohol level (BAL), the amount of alcohol in your blood
stream (here represented by B(t), is measured in milligrams of alcohol per 10millilitres of blood. At the end of a party (time t = 0), a drinker is found to have
B(0) = 0.08 (the legal level for driving impairment), and after that time, B(t)
satisfies the differential equation
dB
= −kB,
dt
k>0
where k is a constant that represents the rate of removal of alcohol form the blood
stream by the liver.
(a) If the drinker had waited for 3 hrs before driving (until = 3), his BAL would
have dropped to 0.04. Determine the value of the rate constant k (specifying
appropriate units) for this drinker.
(b) According to the model, how much longer would it take for the BAL to drop
to 0.01?
0.7. Population with immigration: An island has a bird population of density P (t).
New birds arrive continually with a constant colonization rate C birds per day. Each
bird also has a constant probability per day, γ, of leaving the Island. At time t = 0
the bird population is P (0) = P0
(a) Write down a differential equation that describes the rate of change of the bird
population on the island.
(b) Find the steady state of that equation and interpret this in terms of the bird
population.
c Write down the solution of the differential equation you found in (b) and show
that it satisfies the following two properties: (i) the initial condition, (ii) as
t → ∞ it approaches the steady state you found in (b).
Exercises
v
(d) If the island has no birds on it at time t = 0, how long would it take for the
bird population to grow to 80% of the steady state value?
0.8. Learning and forgetting: Knowledge can be acquired by studying, but it is forgotten over time A simple model for learning represents the amount of knowledge,
y(t), that a person has at time t (in years) by a differential equation
dy
= S − fy
dt
where S ≥ 0 is the rate of studying and f ≥ 0 is the rate of forgetting. We will
assume that S and f are constants that are different for each person. [Your answers
to the following questions will contain constants such as S or f .]
(a) Mary never forgets anything. What does this imply about the constants S and
f ? Mary starts studying in school at time t = 0 with no knowledge at all. How
much knowledge will she have after 4 years (i.e. at t = 4)?
(b) Tom learned so much in preschool that his knowledge when entering school
at time t = 0 is y = 100. However, once Tom in school, he stops studying
completely. What does this imply about the constants S and f ? How long will
it take him to forget 75% of what he knew?
(c) Jane studies at the rate of 10 units per year and forgets at rate of 0.2 per year.
Sketch a “direction field” (“slope field”) for the differential equation describing Jane’s knowledge. Add a few curves y(t) to show how Jane’s knowledge
changes with time.
0.9. Least cost: A rectangular plot of land has dimensions L by D. A pipe is to be built
joining points A and C. The pipe can be above ground along the border of the plot
(Section AB), but has to be buried underground along the segment BC. The cost
per unit length of the underground portion is 3 times that of the cost of the above
ground portion. Determine the distance y so that the cost of the pipe will be as low
as possible.
C
y
L
B
D
Figure 1. Figure for problem 9
A
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Review Problems
0.10. Ducks in a row:
Graduate student Ryan Lukeman studies behaviour of duck flocks swimming near
Canada Place in Vancouver, BC. This figure from his PhD thesis shows his photography set-up. Here H = 10 meters is the height from sea level up to his camera
aperture at the observation point, D = 2 meters is the width of a pier (a stationary
platform whose size is fixed), and x is the distance from the pier to the leading duck
in the flock (in meters). α is a visual angle subtended at the camera, as shown. If
the visual angle is increasing at the rate of 1/100 radians per second, at what rate is
the distance x changing at the instant that x = 3 meters?
α
H
D
x
Figure 2. Figure for problem 10
0.11. Circular race track: Two runners are running around a circular race track whose
length is 400m, as shown in Fig. 3(a). The first runner make a full revolution every
100s and the second runner every 150 s. They start at the same time at the start
position, and the angles subtended by each runner with the radius of the start position
are θ1 (t), θ2 (t), respectively. As the runners go around the track both θ1 (t) and θ2 (t)
will be changing with time.
R1
(a)
R2
START
(b)
START
Figure 3. Figure for problems 11 and 12. The angles in (a) are θ1 (t), θ2 (t). In
(b), the angle between the runners is φ.
Exercises
vii
(a) At what rate it the angle φ = θ1 − θ2 changing?
(b) What is the angle φ at t = 25s?
(c) What is the distance between the runners at t = 25s? (Here “distance” refers
to the length of the straight line connecting the runners.)
(d) At what rate is the distance between the runners changing at t = 25s?
0.12. Phase angle and synchrony: Suppose that the same two runners as in Problem 11
would speed up or slow down depending on the angle between them, φ. (See Fig. 3).
Then φ = φ(t) will change with time. We will assume that the angle φ satisfies a
differential equation of the form
dφ
= A − B sin(φ)
dt
where A, B > 0 are constants.
(a) What values of φ correspond to steady states (i.e. constant solutions) of this
differential equation?
(b) What restriction should be placed on the constants A, B for these steady states
to exist?
(c) Suppose A = 1, B = 2. Sketch the graph of f (φ) = A − B sin(φ) for
−π ≤ φ ≤ π and use it to determine what will happen if the two runners start
st the same point, (φ = 0) at time t = 0.
0.13. Logistic equation and its solution:
(a) Show that the function
y(t) =
1
1 + e−t
satisfies the differential equation
dy
= y(1 − y).
dt
(b) What is the initial value of y at t = 0?
(b) For what value of y is the growth rate largest?
(d) What will happen to y after a very long time?
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Review Problems
0.0.1 Answers to Chapter 14 Problems
• Problem 14.1:
1(d), 2(b), 3(a), 4(c) , 5(a)
• Problem 14.2:
3/4 cm/sec.
• Problem 14.3:
(a) v(t) = 3(t + 2)2 + λ, a(t) = dv/dt = 6(t + 2). (b) λ.
• Problem 14.4:
(a) t = 1 and t = 4, (b) 9 m/s (c) 12 m.
• Problem 14.5:
(a)x = 0, ln(A)/b (b)x = 1/b.
• Problem 14.6:
(a) k = ln(2)/3 per hr (b) 6 more hrs.
• Problem 14.7:
(a) dP/dt = C − γP (b) P = C/γ (d) t = (ln(1/0.2)/γ.
• Problem 14.8:
(a) y = 4S (b) T = 2τ1/2 = 2 ln(2)/f (c) y → 50
• Problem 14.9:
√
y = D/ 8
• Problem 14.10:
0.125m/s
• Problem 14.11:
(a) π/150 radians/s (b) π/6 radians (c) D =
• Problem 14.12:
200
π (2
−
√ 1/2
(d)
3)
(a) φ = arcsin(A/B), (b) −1 < arcsin(A/B) < 1, (c) φ → π/6
• Problem 14.13:
(b) y(0) = 0.5 (c) at y = 0.5 (d) y → 1.
ix
dD
dt
=
2
√
3(2− 3)1/2
x
Short Answers to Problems
0.0.2 Solutions to Chapter 14 Problems
• Solution to Problem 14.1:
1 The equation is obtained as
y − f (x0 )
= f ′ (x0 ),
x − x0
y = f (x0 ) + f ′ (x0 )(x − x0 )
2 The distance between the functions f (x) and g(x) is f (x) − g(x) = x2 − x3 .
Call this h(x). This is the function we wish to maximize.
h(x) = x2 − x3 ,
h′ (x) = 2x − 3x2 ,
h′′ (x) = 2 − 6x
So h′ (x) = 0 when x = 2/3. At that point, h′′ (x) < 0 so this is a local
maximum. The answer is x = 2/3.
3 The equation is x2 − y 2 = 1. Implicit differentiation leads to 2x − 2yy ′ = 0.
rearranging results in y ′ = −2x/(−2y) = x/y. The x coordinate is√x = 2.
The corresponding y coordinate is y 2 = x2 − 1 = 22 − 1 = 3 so√
y = 3 is the
y coordinate of the point. Thus the slope at that point is y ′ = 2/ 3.
4 Given ln(x) = 2 ln(a) − 3 ln(b) we rewrite this as ln(x) = ln(a2 ) − ln(b3 ) =
ln(a2 /b3 so x = a2 /b3
5 We have the following:
y = arctan (x) − x/2,
, y′ =
1
1
− ,
1 + x2
2
y ′′ = −2
x
(1 + x2 )
2
So y ′ = 0 when 1 + x2 = 2, x2 = 1, x = ±1. So y ′′ < 0 for positive x,
and x = 1 is a local maximum, y ′′ changes sign at x = 0 (inflection point) and
y ′′ > 0 for x = −1 (local minimum). The correct answer is LX: x = 1, LM:
x = −1, IP: x = 0.
• Solution to Problem 14.2:
r13 + r23 = constant so r12 r1 (t)′ + r22 r2′ (t) = 0, r2′ (t) = −r12 /r22 r1 (t)′ = 3/4 cm/sec.
• Solution to Problem 14.3:
(a) The velocity is v(t) = dx/dt = 3(t + 2)2 + λ. The acceleration is a(t) =
d2 x/dt2 = dv/dt = 6(t + 2).
(b) The minimal velocity occurs when dv/dt = 0 which implies a = 0, which
happens when t = −2. To check that this is a minimum we compute d2 v/dt2 =
6 > 0 so the velocity function is concave up and the critical point is a local
minimum. We also find that the minimal velocity is v(−2) = 3(−2+2)2 +λ =
λ.
• Solution to Problem 14.4:
The displacement is y(t) = t3 − 6t2 + 9t. The velocity is thus v(t) = s′ (t) =
3t2 − 12t + 9, and the acceleration is a(t) = v ′ (t) = 6t − 12.
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Solutions to Problems
(a) To maximize y(t) we look for time when y ′ = 0, By the above, y ′ (t) = 0 =
3t2 − 12t + 9 = 3(t − 3)(t − 1), so critical points occur at t = 1 and t = 3,
and this is when the direction of motion changes We consider the values at
displacement at the critical points and at the endpoints t = 0 or t = 4: We have
y(0) = 0, y(1) = 4, y(3) = 0, y(4) = 4 so the furthest position is 4m away,
and this occurs at both t = 1 and t = 4.
(b) Critical points of the velocity occur when v ′ (t) = a(t) = 0, i.e. when t = 2,
and v(2) = −3 m/s. However v ′′ (t) = 6 > 0 means that this is a local
minimum. We look at the endpoints: v(0) = v(4) = 9 m/s. So 9 m/s is the
highest velocity.
(c) To get the total distance, we have to consider each portion of the motion between the direction changes, and add up all the contributions. This gives D =
|y(1)−y(0)|+|y(3)−y(1)|+|y(4)−y(3)| = |4−0|+|0−4|+|4−0| = 12m.
• Solution to Problem 14.5:
(a) No change implies y = x so that
x = Axe−bx .
This happens if either x = 0 or if
1 = Ae−bx
⇒
ebx = A
⇒
x=
ln(A)
.
b
(b) To find the number of salmon that lead to the largest level next year, we should
maximize y = Axe−bx with respect to x. We compute
y ′ (x) = A(e−bx − bxe−bx ) = Ae−bx (1 − bx)
Setting y ′ (x) = 0 and noting that the exponential function is always positive
leave us with 1 − bx = 0 so x = 1/b.
• Solution to Problem 14.6:
(a) We are given the differential equation and initial condition
dB
= −kB,
dt
B(0) = 0.08.
At t = 3, the BAL is 0.04, half of the initial level. This implies that the
half-life is 3 hrs, and hence without further calculations, we can conclude that
k = ln(2)/3 per hr.
(b) We are asked how much longer it takes for the BAL to drop to 0.01. This is 1/4
of the level (0.04), i.e. (1/2)2 . Hence it would take two additional half-lives,
or 6 hrs to drop to that level.
• Solution to Problem 14.7:
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(a) The differential equation that describes the rate of change of the bird population
on the island is:
dP
= rate arrival − rate departure = C − γP
dt
We note that every term has the same units, namely birds/day.
(b) The steady state of this equation is obtained from dP/dt = 0 which leads to
P = C/γ birds. At this population level, there will be no change as the arrival
and departure of birds is balanced.
c From our experience with equations of the form dy/dt = a − by we know that
the solution of the bird population equation would be
C
C
e−γt .
+ P0 −
P (t) =
γ
γ
It is easy to see that at t = 0 we have
C
C
C
C
P (0) =
e0 =
= P0
+ P0 −
+ P0 −
γ
γ
γ
γ
so that the initial condition is satisfied (property i), (ii) as t → ∞ we have that
C
C
C
e−γt → .
+ P0 −
e−γt → 0, ⇒ P (t) =
γ
γ
γ
so the population approaches its steady state value with time.
(d) If the island has no birds on it at time t = 0, then P (0) = P0 = 0 We write
C
C
C
e−γt = 0.8
+ 0−
γ
γ
γ
to find when the population has grown to 80% of its steady state value. Now
we solve for t:
C
C −γt
e
= 0.2
γ
γ
Cancelling out the common factors leads to
e−γt = 0.2,
⇒
−γt = ln(0.2)
⇒
t=
ln(1/0.2)
ln(0.2)
=
−γ
γ
• Solution to Problem 14.8:
(a) For Mary, dy
dt = S for a constant S > 0 and y(0) = 0 so after 4 years Mary’s
knowledge level is y = 4S.
(b) For Tom dy
dt = −f y for f > 0 a constant and y(0) = 100. Then y(t) =
100e−f t. His knowledge “half life” is τ1/2 = ln(2)/f so he forgets half within
that time and he forgets half again after another period of this length, so he has
forgotten 1/2 + 1/4 = 3/4 = 0.75 after T = 2τ1/2 = 2 ln(2)/f .
xiv
Solutions to Problems
100
80
60
40
20
0
0
2
4
6
8
10
Figure 4. Figure for solution to problem 8
(c) For Jane, S = 10 units per year, f = 0.2 per year, so dy
dt = 10 − 0.2y. After a
long time dy/dt → 0, y → 10/0.2 = 50. A direction field is shown.
• Solution to Problem 14.9:
The total cost of the pipe (including segments AB and BC is
p
T (y) = (L − y) + 3 D2 + y 2
Differentiating with respect to y and setting the derivative to zero to find a critical
point leads to
y
dT
=0
= −1 + 3 p
2
dy
D + y2
⇒
y
3p
=1
2
D + y2
p
After algebraic simplification,
we get 3y = D2 + y 2 or 9y 2 = D2 +y 2 so, solving
√
for y leads to y = D/ 8. We ask if this is a minimum or maximum.
Using the first derivative test, we see that dT /dy = −1 when y = 0, then dT /dy = 0
at the critical point and for larger y, dT /dy > 0. Hence the pattern of the first
derivative verifies that the critical point is a local minimum.
We can also note that
p the endpoints of the interval, namly y = 0, L lead to T =
L + 3D and T = 3 D2 + y 2 , respectively.
• Solution to Problem 14.10:
In the figure, the angle θ shown does not change. Only α is changing.
tan(α + θ) =
D+x
H
sec2 (α + θ)α′ (t) =
1 ′
x (t)
H
x′ (t) = H sec2 (α + θ)α′ (t)
xv
In the large triangle, the hypotenuse is
p
H 2 + (D + x)2 , so
sec2 (α + θ) =
H 2 + (D + x)2
H2
Thus
x′ (t) = H
H 2 + (D + x)2 ′
H 2 + (D + x)2 ′
α
(t)
=
α (t)
H2
H
Using H = 10, D = 2, x = 3, α′ (t) = 1/100 we get
x′ (t) =
100 + 25 1
= 0.125m/s
10
100
Alternately, this can be done using inverse trigonometric functions:
D+x
H
1
α (t) =
1 + [(D + x)/H]2
α = arctan
′
−θ
1
H
x′ (t)
• Solution to Problem 14.11:
(a) Let θ1 , θ2 be the angles made by runners 1 and 2. Then
2π
dθ1
=
,
dt
100
dθ1
2π
=
dt
150
For φ = θ1 − θ2 the angle between the runners we have that
dπ
dθ1
dθ2
=
−
= 2π
dt
dt
dt
1
1
−
100 150
=
π
.
150
(b) When t = 25 runner 1 made 1/4 of a revolution and is thus at θ1 = π/2. Runner
2 made 1/6 of a revolution and is at θ2 = π/3. The angle between them is then
φ = (π/2) − (π/3) = π/6.
(c) The distance between the runners can be found using the law of cosines
√
D2 = 2r2 − 2r2 cos(φ) = 2r2 (1 − cos(φ)) ⇒ D = 2r(1 − cos(φ))1/2
The radius of the track is r = 400/2π = 200/π so at t = 25s the distance is
√ 200
√ 200
D= 2
(1 − cos(π/6))1/2 = 2
π
π
√ !1/2
√
3
200
1−
=
(2 − 3)1/2 .
2
π
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Solutions to Problems
(d) The rate of change of the distance can be found by differentiating the law of
cosines with respect to time
D2 = 2r2 − 2r2 cos(φ)
dφ
dD
= 2r2 sin(φ)
2D
dt
dt
dD
r2
dφ
=
sin(φ)
dt
D
dt
(1)
Using φ = π/6 and the value of D from part (c), we get
π
1
dD
200 1
2
√
√
=
=
dt
π 2 (2 − 3)1/2 150
3(2 − 3)1/2
• Solution to Problem 14.12:
(a) Steady states satisfy
dφ
= A − B sin(φ) = 0
dt
⇒
sin(φ) =
A
B
⇒
φ = arcsin(A/B).
(b) The above steady states exist only if −1 < arcsin(A/B) < 1.
(c) Starting from phase φ = 0 the points will eventually approach the steady state
shown by the dark dot in Fig 5. Since A/B = 1/2, this point will satisfy
sin(φ) = 1/2 so that φ = π/6. The runners will be at phase difference π/6
after a while.
A/B
o
Figure 5. Figure for solution to problem 12.
• Solution to Problem 14.13:
(a) We compute the derivative of the function and find:
e−t
1
dy(t)
(−e−t ) =
=−
−t
2
dt
(1 − e )
(1 − e−t )2
xvii
We also have that
y(t) =
1
,
1 + e−t
1 − y(t) =
e−t
,
1 + e−t
y(1 − y) =
e−t
.
(1 − e−t )2
Thus, we find that both sides are equal, so that the given function satisfies the
differential equation.
(b) We find that y(0) =
1
1+e0
=
1
2
= 0.5.
(b) The growth rate is f (y) = y(1 − y). this is a parabola opening downwards with
a vertex at y = 0.5 which is also a local maximum. Thus the growth rate is
largest when y = 0.5.
(d) After a long time, y → 1 which is a stable steady state in this case.